Friction and initial velocity

[HCHopeful]

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I did a quick search, but my question is rather narrow.

Okay, I have a coefficient of friction of 0.5. I was traveling down the highway at an unknown velocity and slammed on my breaks just because. I hopped out and found my skid marks to be 10 m long. What was my initial velocity?

I'm awful with friction apparently, and this is something I really am not understanding. Thanks everyone.

 

[Cawolf]

Work = Force x distance = F(friction) x 10 = 0.5mg x 10 = 5mg

Work = change in energy = final KE - initial KE = 0 - KE = -KE = -.5mv^2 (we will use the absolute value because the negative denotes direction and we are looking at work which is scalar)

S0 now we can set the two equations for work equal to each other.

5mg = 0.5mv^2

5g = 0.5v^2

v^2 = 10g

v = sqrt (98)

v = 9.9 m/s

 

[HCHopeful]

Work = Force x distance = F(friction) x 10 = 0.5mg x 10 = 5mg

Work = change in energy = final KE - initial KE = 0 - KE = -KE = -.5mv^2 (we will use the absolute value because the negative denotes direction and we are looking at work which is scalar)

S0 now we can set the two equations for work equal to each other.

5mg = 0.5mv^2

5g = 0.5v^2

v^2 = 10g

v = sqrt (980)

v = 31 m/s

Thank you very much. I appreciate it!

 

[HCHopeful]

Work = Force x distance = F(friction) x 10 = 0.5mg x 10 = 5mg

Work = change in energy = final KE - initial KE = 0 - KE = -KE = -.5mv^2 (we will use the absolute value because the negative denotes direction and we are looking at work which is scalar)

S0 now we can set the two equations for work equal to each other.

5mg = 0.5mv^2

5g = 0.5v^2

v^2 = 10g

v = sqrt (980)

v = 31 m/s

Whoops, just caught this. It appears you meant "v = sqrt (98)" which would give a value approximately equal to 10 m/s.

 

[Cawolf]

Yah sorry you are correct.

Other than that you see how it is just an application of the work-energy theorem and the mass is irrelevant because it cancels.

 

[HCHopeful]

Yah sorry you are correct.

Other than that you see how it is just an application of the work-energy theorem and the mass is irrelevant because it cancels.

Yeah, I should have caught on to that pretty quickly. Thanks, again!